Logarithm Basics Part III

The given equation can be expressed as

$\sqrt{(1+2\log_2x)(2+\frac{\log_2x}{2})}=\frac{3}{2}\log_2x$

Let $\log_2x=m$ ,then

$\sqrt{(1+2m)(2+\frac{m} {2})}=\frac{3}{2}m$

$(1+2m)(2+\frac{m} {2})=\frac{9}{4}m^2$

$2+\frac{9}{2}m=\frac{5}{4}m^2$

$5m^2-18m-8=0$

$(m-4)(5m+2)=0$

$m=4,-2/5$ $\implies$ $x=16,2^{-2/5}$

We reject $2^{-2/5}$ $\therefore$ $x=\boxed{16}$

Thus the answer is $\dfrac{16}{1}=\boxed{16}$

$\square$

Similar method as Parth Lohomi but then I initially accepted both roots. Then I just guessed that the non-integral root should be rejected then presto! Anyway thank you Sandeep Bhardwaj for enlightening me on why that root should be rejected!! :)

Can someone tell me why did we have to ignore the (0.5)^0.4 part? Sandeep Bhardwaj sir